Where's Forever?

$1+x+x^2 +x^3+x^4+... = 15$ $x(1+x+x^2+x^3+x^4+...) =14$ $x(1+(x+x^2+x^3+x^4+...)) =14$ $x(1+14)=14$

$\boxed{x=\frac{14}{15}}$

First note that this is an infinite geometric series so | ${x}$ | < 1, so ${x}$ = $\frac{15}{14}$ is clearly not the answer. Now: $1+x + x^2 + x^3 + x^4 + .... = \frac{1}{1-x} = 15 \Rightarrow\ x= 1-\frac{1}{15}$ so $x = \frac{14}{15}$

as x+x^2+x^3+x^4...=14.....(i) take x common so it will become 1+x+x^2+x^3+x^4...=14/x...(ii) also we can write the equation (ii) as 1+(x+x^2+x^3+x^4...)=14/x...(iii) substituting equation (i) in equation (iii) will become 1+14=14/x ==> x=14/15

we know that x + x^2 + x^3........=14 Now, take x common. Then x(1 + x + x^2 + x^3 + x^4.....) = 14 x(1+14)=14 x(15) = 14 so, x = 14/15

It is a geometric series with sum x/(1-x) for x between 0 and 1.

Then the equation can be written as: x/(1-x) = 14

From which follows that x = 14/15

Qed